A force of `(3hati-1.5hatj)N` acts on `5kg` body. The body is at a position of `(2hati-3hatj)m` and is travelling at `4m//s`. The force acts on the body until it is at the position `(hati+5hatj)m`. Calculate final.speed.

Options

A. `sqrt(20)m//s`
B. `sqrt(10)m//s`
C. `sqrt(40)m//s`
D. `sqrt(50)m//s`

Explanation

To solve the problem step by step, we will follow the physics principles of force, acceleration, displacement, work done, and kinetic energy.

### Step 1: Identify the given values
- Force \( \vec{F} = 3 \hat{i} - 1.5 \hat{j} \, \text{N} \)
- Mass \( m = 5 \, \text{kg} \)
- Initial position \( \vec{r}_i = 2 \hat{i} - 3 \hat{j} \, \text{m} \)
- Final position \( \vec{r}_f = \hat{i} + 5 \hat{j} \, \text{m} \)
- Initial velocity \( \vec{v}_i = 4 \, \text{m/s} \)

### Step 2: Calculate the acceleration
Using Newton's second law, the acceleration \( \vec{a} \) can be calculated as:
\[ \vec{a} = \frac{\vec{F}}{m} = \frac{3 \hat{i} - 1.5 \hat{j}}{5} = \left( \frac{3}{5} \hat{i} - \frac{1.5}{5} \hat{j} \right) = 0.6 \hat{i} - 0.3 \hat{j} \, \text{m/s}^2 \]

### Step 3: Calculate the displacement
The displacement \( \vec{S} \) is given by:
\[ \vec{S} = \vec{r}_f - \vec{r}_i = (\hat{i} + 5 \hat{j}) - (2 \hat{i} - 3 \hat{j}) = (-1 \hat{i} + 8 \hat{j}) \, \text{m} \]

### Step 4: Calculate the work done
Work done \( W \) is calculated using the dot product of force and displacement:
\[ W = \vec{F} \cdot \vec{S} = (3 \hat{i} - 1.5 \hat{j}) \cdot (-1 \hat{i} + 8 \hat{j}) \]
Calculating the dot product:
\[ W = 3 \cdot (-1) + (-1.5) \cdot 8 = -3 - 12 = -15 \, \text{J} \]

### Step 5: Relate work done to change in kinetic energy
The work-energy theorem states that the work done is equal to the change in kinetic energy:
\[ W = \Delta KE = KE_f - KE_i \]
Where:
\[ KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \cdot 5 \cdot (4^2) = \frac{1}{2} \cdot 5 \cdot 16 = 40 \, \text{J} \]

### Step 6: Set up the equation for final kinetic energy
Let \( v_f \) be the final speed. Then:
\[ W = \frac{1}{2} m v_f^2 - KE_i \]
Substituting the values:
\[ -15 = \frac{1}{2} \cdot 5 v_f^2 - 40 \]

### Step 7: Solve for \( v_f^2 \)
Rearranging the equation gives:
\[ \frac{1}{2} \cdot 5 v_f^2 = 40 - 15 \]
\[ \frac{5}{2} v_f^2 = 25 \]
\[ v_f^2 = \frac{25 \cdot 2}{5} = 10 \]

### Step 8: Calculate the final speed
Taking the square root:
\[ v_f = \sqrt{10} \, \text{m/s} \]

### Final Answer
The final speed of the body is \( \sqrt{10} \, \text{m/s} \).

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